Wednesday, December 5, 2007
Monday, December 3, 2007
review session for final
Sophia will hold a review session for the final on Monday, December 10 the same time and place as the usual class.
Problem 6d on PS8
Instead of what is currently in the problem, consider the following expression for B2 in terms of the configurational integrals,
B2(T) = (Z1^2 - Z2) / (2 Z1^2 / V)
What was written in the problem statement is a simplification of this expression for the case in which there are only pair interactions in the system (no fields).
B2(T) = (Z1^2 - Z2) / (2 Z1^2 / V)
What was written in the problem statement is a simplification of this expression for the case in which there are only pair interactions in the system (no fields).
Thursday, November 29, 2007
Negative temperature
Let's think about the concept of temperature for a moment. What is a temperature, in terms of how you measure it in real life? When we measure temperature, we put a small system (a thermometer) in contact with a larger system. The thermometer comes to equilibrium with the larger system by exchanging energy until the total entropy of both systems is maximized. What we read off as temperature has to do with how the fluid inside the thermometer expands or contracts.
Systems at negative temperature come to equilibrium by the same means. Recall that, when we put two systems together that are isolated, the total entropy is maximized, S = S1+S2 while the total energy is held constant, E=E1+E2. Thus the sign of dS must be positive. But dS is just related to the temperature, when the energy changes.
A thermometer put in contact with a system at negative temperature will come to equilibrium in the same way it would as if it were put into contact with a system at positive temperature. The two systems will exchange energy until their temperatures are the same. Whatever temperature the thermometer is at in the end will depend on how much energy it exchanged with the system.
Not all systems can have negative temperatures. Only systems that have entropy functions that contain regions of negative slope do. Some systems have entropy functions that increase monotonically with energy for all energies, and thus, these systems cannot have negative slopes. Fluids have the latter kind of entropy function. Magnets can have the former. The basic difference between the two is that fluids can have an infinite energy (because the momenta are unbounded), but magnets can be saturated at a maximum energy.
Systems at negative temperature come to equilibrium by the same means. Recall that, when we put two systems together that are isolated, the total entropy is maximized, S = S1+S2 while the total energy is held constant, E=E1+E2. Thus the sign of dS must be positive. But dS is just related to the temperature, when the energy changes.
A thermometer put in contact with a system at negative temperature will come to equilibrium in the same way it would as if it were put into contact with a system at positive temperature. The two systems will exchange energy until their temperatures are the same. Whatever temperature the thermometer is at in the end will depend on how much energy it exchanged with the system.
Not all systems can have negative temperatures. Only systems that have entropy functions that contain regions of negative slope do. Some systems have entropy functions that increase monotonically with energy for all energies, and thus, these systems cannot have negative slopes. Fluids have the latter kind of entropy function. Magnets can have the former. The basic difference between the two is that fluids can have an infinite energy (because the momenta are unbounded), but magnets can be saturated at a maximum energy.
Tuesday, November 27, 2007
Monday, November 19, 2007
Correction to problem 6d on PS6
It turns out that neither bottle can be frozen the way mentioned. Here is a new problem statement for 6d:
An alternative way to freeze the solutions is to change the pressure. At 0 °C for each bottle, what pressure must be reached in order for freezing to begin? Assume the water freezes first, and assume constant molar densities. Why are these numbers unrealistic? What is special about water freezing here, relative to other liquids?
An alternative way to freeze the solutions is to change the pressure. At 0 °C for each bottle, what pressure must be reached in order for freezing to begin? Assume the water freezes first, and assume constant molar densities. Why are these numbers unrealistic? What is special about water freezing here, relative to other liquids?
Hints for problem 5 on PS6
If you are having trouble with problem 5, go back and take a look at the lecture "Stability" to see a related derivation of tangent line properties, for the liquid-vapor transition.
Also, don't forget that the conditions for the critical point for the liquid-vapor transition were:
d^2 a /dv^2 = 0 and d^3 a / dv^3 = 0
A similar set of two equations exists for the mixture.
Also, don't forget that the conditions for the critical point for the liquid-vapor transition were:
d^2 a /dv^2 = 0 and d^3 a / dv^3 = 0
A similar set of two equations exists for the mixture.
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